X Yi 3
(x + iy)
From Pascal's triangle:
(a + b) 3 = a 3 + 3a 2 * b + 3ab 2 + b 3.
Then:
(x + iy) 3
= x 3 + 3 (x) 2 (iy) +3 (x) (iy) 2 + (iy) 3
= x 3 + (3x 2 * y) i 3xy 2 (y 3) i
= (x 3 3xy 2) + (3x 2 * y y 3) i.
It helps!
Note: Usually follows the coefficient and i 2 = 1.
(x + yi) 3 = (x + yi) (x + yi) (x + yi) = (x 2 + xyi + xyi + yi 2) (x + yi) = (x 2 + 2xyi y ) (x + yi) =
(x 3 + x 2yi + 2x 2yi + 2xy 2i 2xy y 2i) =
(x 3 + 3x 2yi + 2xy 2xy y 2i)
(Answer)> x 3 2xy 2xy + 3x 2yi y 2i
(x + iy) 3
(x + iy) (x + iy) (x + iy)
X 2xy + y 2 (x + iy)
x 3 + x 2y + 2y 2 + iXyxy 2i + iy 3
x 3 + x 2y + 2y 2 + iXyxy 2i + iy 3
